Answer
\[9\ln \left| x \right| + \frac{{15}}{2}{e^{ - 0.4x}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {\frac{9}{x} - 3{e^{ - 0.4x}}} \right)dx} \hfill \\
By\,\,the\,\,sum\,\,and\,\,differentiate\,\,rule \hfill \\
\int_{}^{} {\frac{9}{x}dx - \int_{}^{} {3{e^{ - 0.4x}}dx} } \hfill \\
9\int_{}^{} {{x^{ - 1}}dx} - 3\int_{}^{} {{e^{ - 0.4x}}dx} \hfill \\
Use\,\,indefinite\,\,integrals \hfill \\
\int_{}^{} {{x^{ - 1}}dx = \ln \left| x \right| + C\,\,,\,\,\,\int_{}^{} {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C} \hfill \\
9\ln \left| x \right| - \frac{{3{e^{ - 0.4x}}}}{{ - 0.4}} + C \hfill \\
Simplifying \hfill \\
9\ln \left| x \right| + \frac{{15}}{2}{e^{ - 0.4x}} + C \hfill \\
\end{gathered} \]
